Question: $f(n) = 6n^{3}+2n^{2}$ $g(t) = -5t^{2}+6t+4(f(t))$ $ g(f(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 6(0^{3})+2(0^{2})$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = -5(0^{2})+(6)(0)+4(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 6(0^{3})+2(0^{2})$ $f(0) = 0$ That means $g(0) = -5(0^{2})+(6)(0)+(4)(0)$ $g(0) = 0$